Mathematics - Limits Question with Solution | TestHub

Let, $$\begin{array}{rl}& L=\underset{x\to 0}{\lim }\frac{(x\tan 2x-2x\tan x)}{(1-\cos 2x{)}^2}=\underset{x\to 0}{\lim }K\text{ (say) }\\ & \Rightarrow K=\frac{x\left[\frac{2\tan x}{1-(\tan x{)}^2}\right]-2x\tan x}{{\left(1-\left(1-2{\sin }^{2}x\right)\right)}^2}\\ & =\frac{2x\tan x-\left[2x\tan x-2x{\tan }^{3}x\right]}{4{\sin }^{4}x\times \left(1-{\tan }^{2}x\right)}\\ & =\frac{2x{\tan }^{3}x}{4{\sin }^{4}x\times \left(1-{\tan }^{2}x\right)}\\ & =\frac{2x{\tan }^{3}x}{4{\sin }^{4}x\times \left(\frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x}\right)}\end{array}$$ $$\begin{array}{rl}& =\frac{2x\frac{{\sin }^{3}x}{{\cos }^{3}x}}{4{\sin }^{4}x\times \left(\frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x}\right)}\\ & \Rightarrow K=\frac{x}{2\sin x\times \left({\cos }^{2}x-{\sin }^{2}x\right)\cos x}\\ \therefore L& =\underset{x\to 0}{\lim }\frac{x}{2\sin x}\times \underset{x\to 0}{\lim }\frac{1}{\cos x\left({\cos }^{2}x-{\sin }^{2}x\right)}\\ & =\underset{x\to 0}{\lim }\frac{x}{2\sin x}\times \underset{x\to 0}{\lim }\frac{1}{\cos 0\left({\cos }^{2}0-{\sin }^{2}0\right)}\\ & =\frac{1}{2}\end{array}$$